Find $\int \dfrac{1}{5x^2-20x+100}\,dx$. Choose 1 answer: Choose 1 answer: (Choice A) A $\text{arctan}\left(\dfrac{x-2}{4}\right)+C$ (Choice B) B $\dfrac{1}{20} \text{arcsin}\left(\dfrac{x-2}{4}\right)+C$ (Choice C) C $\text{arcsin}\left(\dfrac{x-2}{4}\right)+C$ (Choice D) D $\dfrac{1}{20} \text{arctan}\left(\dfrac{x-2}{4}\right)+C$
Explanation: The integrand is in the form $\dfrac{1}{p(x)}$ where $p(x)$ is a quadratic expression. This suggests that we should rewrite $p(x)$ by completing the square. Specifically, we will rewrite $p(x)$ as a constant multiple of $(x+h)^2+k^2$. Then, we will be able to integrate using the following formula, which is based on the derivative of the inverse tangent function: $\int \dfrac{1}{(x+ h)^2+ k^2}\,dx=\dfrac{1}{ k} \text{arctan}\left(\dfrac{x+ h}{ k}\right)+C$ [Why is this formula true?] We start by rewriting $p(x)$ as a constant multiple of $(x+ h)^2+ k^2$ : $\begin{aligned} 5x^2-20x+100&=5[x^2-4x+20] \\\\ &=5[x^2-4x+4+16] \\\\ &=5[(x-2)^2+16] \\\\ &=5[(x{-2})^2+{4}^2] \end{aligned}$ Now we can find the integral: $\begin{aligned} &\phantom{=}\int \dfrac{1}{5x^2-20x+100}\,dx \\\\ &=\int\dfrac{1}{5[(x-2)^2+4^2]}\,dx \\\\ &=\dfrac15\int\dfrac{1}{(x{-2})^2+{4}^2}\,dx \\\\ &=\dfrac15\cdot\dfrac{1}{{4}} \text{arctan}\left(\dfrac{x{-2}}{{4}}\right)+C \\\\ &=\dfrac{1}{20} \text{arctan}\left(\dfrac{x-2}{4}\right)+C \end{aligned}$ In conclusion, $\int \dfrac{1}{5x^2-20x+100}\,dx=\dfrac{1}{20} \text{arctan}\left(\dfrac{x-2}{4}\right)+C$